A large box of mass 4kg slides across a frictionless surface with a velocity of 12m/s to the right. It collides with a smaller box of mass 2kg that is stationary. The boxes stick together. What is the velocity of the two combined masses after the collision.

Explanation:

Let's call the large box "1" and the smaller box "2". The equation for this works so that momentum is conserved, as the Law states.

That says that the momentum of the larger box plus the momentum of the smaller box before the collision has to equal the momentum of the combined masses since they stick together. Filling in:

[(4(12) + 2(0)] = [(4 + 2)v] and

48 + 0 = 6v so

v = 8 m/s to the right

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: